{"id":252066,"date":"2025-07-11T06:31:47","date_gmt":"2025-07-11T06:31:47","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=252066"},"modified":"2025-07-11T06:31:51","modified_gmt":"2025-07-11T06:31:51","slug":"solving-quadratics-worksheet","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/11\/solving-quadratics-worksheet\/","title":{"rendered":"SOLVING QUADRATICS WORKSHEET"},"content":{"rendered":"\n

SOLVING QUADRATICS WORKSHEET Solve using the Square Root Method 1) -5x – 4 =-504 2) x = 81 Solve by Factoring: 3) 5x” 16x + 3 = 0 4) x + 5x+ 6 = 0 5) x – 16 = 0 6) x _2x =-1 Solve by Completing the Square: 7) 3x? + 18x + 10 = 0 8) x – 2x – 15 = 0 Solve using the Quadratic Formula: 9) 4x” -X-5=0 10) 21″ + 3x – 3 =0<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let’s break down each problem and solve them one by one.<\/p>\n\n\n\n

\n\n\n\n

Square Root Method<\/strong><\/h3>\n\n\n\n

1) -5x – 4 = -504<\/strong><\/p>\n\n\n\n

Step 1: Isolate the term with the variable. \u22125x=\u2212504+4\u2005\u200a\u27f9\u2005\u200a\u22125x=\u2212500-5x = -504 + 4 \\implies -5x = -500\u22125x=\u2212504+4\u27f9\u22125x=\u2212500<\/p>\n\n\n\n

Step 2: Divide by -5. x=\u2212500\u22125\u2005\u200a\u27f9\u2005\u200ax=100x = \\frac{-500}{-5} \\implies x = 100x=\u22125\u2212500\u200b\u27f9x=100<\/p>\n\n\n\n

Answer: x=100x = 100x=100<\/strong><\/p>\n\n\n\n

2) x = 81<\/strong><\/p>\n\n\n\n

This is already in the form x=81x = 81x=81.<\/p>\n\n\n\n

Answer: x=81x = 81x=81<\/strong><\/p>\n\n\n\n

\n\n\n\n

Factoring<\/strong><\/h3>\n\n\n\n

3) 5x\u00b2 – 16x + 3 = 0<\/strong><\/p>\n\n\n\n

To factor, we look for two numbers that multiply to 5\u00d73=155 \\times 3 = 155\u00d73=15 and add to \u221216-16\u221216.<\/p>\n\n\n\n

The pair of numbers that work is -15 and -1. So: 5×2\u221215x\u2212x+3=05x\u00b2 – 15x – x + 3 = 05×2\u221215x\u2212x+3=0<\/p>\n\n\n\n

Now factor by grouping: 5x(x\u22123)\u22121(x\u22123)=05x(x – 3) – 1(x – 3) = 05x(x\u22123)\u22121(x\u22123)=0<\/p>\n\n\n\n

Factor out (x\u22123)(x – 3)(x\u22123): (5x\u22121)(x\u22123)=0(5x – 1)(x – 3) = 0(5x\u22121)(x\u22123)=0<\/p>\n\n\n\n

Set each factor equal to zero: 5x\u22121=0orx\u22123=05x – 1 = 0 \\quad \\text{or} \\quad x – 3 = 05x\u22121=0orx\u22123=0<\/p>\n\n\n\n

Solving each: x=15orx=3x = \\frac{1}{5} \\quad \\text{or} \\quad x = 3x=51\u200borx=3<\/p>\n\n\n\n

Answer: x=15x = \\frac{1}{5}x=51\u200b or x=3x = 3x=3<\/strong><\/p>\n\n\n\n

4) x\u00b2 + 5x + 6 = 0<\/strong><\/p>\n\n\n\n

Look for two numbers that multiply to 6 and add to 5. These are 2 and 3. So: (x+2)(x+3)=0(x + 2)(x + 3) = 0(x+2)(x+3)=0<\/p>\n\n\n\n

Set each factor equal to zero: x+2=0orx+3=0x + 2 = 0 \\quad \\text{or} \\quad x + 3 = 0x+2=0orx+3=0<\/p>\n\n\n\n

Solving each: x=\u22122orx=\u22123x = -2 \\quad \\text{or} \\quad x = -3x=\u22122orx=\u22123<\/p>\n\n\n\n

Answer: x=\u22122x = -2x=\u22122 or x=\u22123x = -3x=\u22123<\/strong><\/p>\n\n\n\n

5) x – 16 = 0<\/strong><\/p>\n\n\n\n

This is a simple linear equation: x=16x = 16x=16<\/p>\n\n\n\n

Answer: x=16x = 16x=16<\/strong><\/p>\n\n\n\n

6) x\u00b2 – 2x = -1<\/strong><\/p>\n\n\n\n

Move the constant to the right side: x2\u22122x+1=0x\u00b2 – 2x + 1 = 0x2\u22122x+1=0<\/p>\n\n\n\n

Factor: (x\u22121)(x\u22121)=0(x – 1)(x – 1) = 0(x\u22121)(x\u22121)=0<\/p>\n\n\n\n

Solve: x=1x = 1x=1<\/p>\n\n\n\n

Answer: x=1x = 1x=1<\/strong><\/p>\n\n\n\n

\n\n\n\n

Completing the Square<\/strong><\/h3>\n\n\n\n

7) 3x\u00b2 + 18x + 10 = 0<\/strong><\/p>\n\n\n\n

Step 1: Move the constant to the other side: 3×2+18x=\u2212103x\u00b2 + 18x = -103×2+18x=\u221210<\/p>\n\n\n\n

Step 2: Divide by 3: x2+6x=\u2212103x\u00b2 + 6x = -\\frac{10}{3}x2+6x=\u2212310\u200b<\/p>\n\n\n\n

Step 3: Take half of 6, square it: (62)2=9\\left(\\frac{6}{2}\\right)\u00b2 = 9(26\u200b)2=9, and add 9 to both sides: x2+6x+9=\u2212103+9x\u00b2 + 6x + 9 = -\\frac{10}{3} + 9×2+6x+9=\u2212310\u200b+9 x2+6x+9=173x\u00b2 + 6x + 9 = \\frac{17}{3}x2+6x+9=317\u200b<\/p>\n\n\n\n

Step 4: Factor the left side: (x+3)2=173(x + 3)\u00b2 = \\frac{17}{3}(x+3)2=317\u200b<\/p>\n\n\n\n

Step 5: Take the square root of both sides: x+3=\u00b1173x + 3 = \\pm \\sqrt{\\frac{17}{3}}x+3=\u00b1317\u200b\u200b<\/p>\n\n\n\n

Step 6: Solve for xxx: x=\u22123\u00b1513x = -3 \\pm \\frac{\\sqrt{51}}{3}x=\u22123\u00b1351\u200b\u200b<\/p>\n\n\n\n

Answer: x=\u22123\u00b1513x = -3 \\pm \\frac{\\sqrt{51}}{3}x=\u22123\u00b1351\u200b\u200b<\/strong><\/p>\n\n\n\n

8) x\u00b2 – 2x – 15 = 0<\/strong><\/p>\n\n\n\n

Step 1: Move the constant to the other side: x2\u22122x=15x\u00b2 – 2x = 15×2\u22122x=15<\/p>\n\n\n\n

Step 2: Take half of -2, square it: (\u221222)2=1\\left(\\frac{-2}{2}\\right)\u00b2 = 1(2\u22122\u200b)2=1, and add 1 to both sides: x2\u22122x+1=15+1x\u00b2 – 2x + 1 = 15 + 1×2\u22122x+1=15+1 (x\u22121)2=16(x – 1)\u00b2 = 16(x\u22121)2=16<\/p>\n\n\n\n

Step 3: Take the square root of both sides: x\u22121=\u00b14x – 1 = \\pm 4x\u22121=\u00b14<\/p>\n\n\n\n

Step 4: Solve for xxx: x=1\u00b14x = 1 \\pm 4x=1\u00b14<\/p>\n\n\n\n

So, x=5x = 5x=5 or x=\u22123x = -3x=\u22123.<\/p>\n\n\n\n

Answer: x=5x = 5x=5 or x=\u22123x = -3x=\u22123<\/strong><\/p>\n\n\n\n

\n\n\n\n

Quadratic Formula<\/strong><\/h3>\n\n\n\n

9) 4x\u00b2 – x – 5 = 0<\/strong><\/p>\n\n\n\n

Use the quadratic formula: x=\u2212(\u22121)\u00b1(\u22121)2\u22124(4)(\u22125)2(4)x = \\frac{-(-1) \\pm \\sqrt{(-1)\u00b2 – 4(4)(-5)}}{2(4)}x=2(4)\u2212(\u22121)\u00b1(\u22121)2\u22124(4)(\u22125)\u200b\u200b x=1\u00b11+808x = \\frac{1 \\pm \\sqrt{1 + 80}}{8}x=81\u00b11+80\u200b\u200b x=1\u00b1818x = \\frac{1 \\pm \\sqrt{81}}{8}x=81\u00b181\u200b\u200b x=1\u00b198x = \\frac{1 \\pm 9}{8}x=81\u00b19\u200b<\/p>\n\n\n\n

So, x=1+98=108=54x = \\frac{1 + 9}{8} = \\frac{10}{8} = \\frac{5}{4}x=81+9\u200b=810\u200b=45\u200b or x=1\u221298=\u221288=\u22121x = \\frac{1 – 9}{8} = \\frac{-8}{8} = -1x=81\u22129\u200b=8\u22128\u200b=\u22121.<\/p>\n\n\n\n

Answer: x=54x = \\frac{5}{4}x=45\u200b or x=\u22121x = -1x=\u22121<\/strong><\/p>\n\n\n\n

10) 2x\u00b2 + 3x – 3 = 0<\/strong><\/p>\n\n\n\n

Use the quadratic formula: x=\u22123\u00b132\u22124(2)(\u22123)2(2)x = \\frac{-3 \\pm \\sqrt{3\u00b2 – 4(2)(-3)}}{2(2)}x=2(2)\u22123\u00b132\u22124(2)(\u22123)\u200b\u200b x=\u22123\u00b19+244x = \\frac{-3 \\pm \\sqrt{9 + 24}}{4}x=4\u22123\u00b19+24\u200b\u200b x=\u22123\u00b1334x = \\frac{-3 \\pm \\sqrt{33}}{4}x=4\u22123\u00b133\u200b\u200b<\/p>\n\n\n\n

Answer: x=\u22123\u00b1334x = \\frac{-3 \\pm \\sqrt{33}}{4}x=4\u22123\u00b133\u200b\u200b<\/strong><\/p>\n\n\n\n

\n\n\n\n

These are the solutions for each equation, along with the necessary steps for each method. Let me know if you need further explanations or if you’d like more examples!<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

SOLVING QUADRATICS WORKSHEET Solve using the Square Root Method 1) -5x – 4 =-504 2) x = 81 Solve by Factoring: 3) 5x” 16x + 3 = 0 4) x + 5x+ 6 = 0 5) x – 16 = 0 6) x _2x =-1 Solve by Completing the Square: 7) 3x? + 18x + […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-252066","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/252066","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=252066"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/252066\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=252066"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=252066"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=252066"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}